Integrate the field between concentric spheres
You will be able to: Derive spherical capacitance from a radial field and a potential integral.
How do two nested spherical conductors store charge?
Place a small conducting sphere inside a concentric conducting shell. Equal opposite charges create a radial field in the gap. The field is stronger near the smaller sphere, so voltage is not a single field value multiplied by gap width.
A useful starting point: A uniform plate field bends a charged particle’s path →
Words and symbols before equations
- Inner radius a
- Radius of the positively charged inner conductor.
- Outer radius b
- Inner radius of the negative enclosing conductor, with b > a.
- Radial integration
- Adding field contributions over successive radial distances dr.
- Spherical capacitance
- Charge divided by the voltage between the two conductors.
What this picture assumes
Vacuum gap between concentric spheres with equal opposite charges. Outer radius b refers to the shell inner surface. Cross-section radii share a scale; field graph covers only the gap, not the metal.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- C = 22.24 pF; inner-to-outer voltage = 44.96 V. Field just outside inner conductor = 899.2 N/C.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For a < r < b, Gauss’s law gives Eᵣ = Q/(4πε₀r²). The field is radial and decreases with distance. Metal on either side has zero field.
The positive inner-to-outer voltage is V(a) − V(b) = ∫ₐᵇEᵣ dr = Q(1/a − 1/b)/(4πε₀). The sign is positive because potential falls along the outward field.
Divide Q by that voltage: C = 4πε₀ab/(b−a). As b becomes extremely large, this tends to the isolated-sphere capacitance 4πε₀a. This is a vacuum, concentric geometry; do not apply the result to off-center spheres.
A worked example, step by step
A spherical capacitor has a = 0.10 m and b = 0.20 m. Estimate C and the voltage when Q = 1 nC.
- Evaluate ab/(b−a) = 0.020/0.10 = 0.20 m.
- C = 4π(8.85×10⁻¹²)(0.20) ≈ 22.24 pF.
- ΔV = Q/C ≈ 44.96 V.
- The gap field varies as 1/r²; using a constant field across the gap would be inaccurate.
The outer radius is the shell’s inner surface radius. Use an integral because E is not uniform.
Does moving the outer shell farther away raise C?
Compare with an explanation
No. It lowers C toward 4πε₀a while the voltage for fixed Q increases.
Predict. Change one thing. Explain.
Vary the radius ratio b/a while holding a and Q fixed. Track capacitance, voltage and the shape of E(r). Compare with the isolated-sphere limit.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
C = 22.24 pF; inner-to-outer voltage = 44.96 V. Field just outside inner conductor = 899.2 N/C.
Vacuum gap between concentric spheres with equal opposite charges. Outer radius b refers to the shell inner surface. Cross-section radii share a scale; field graph covers only the gap, not the metal.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant conductor equilibrium, charge conservation, capacitance or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionDerive the capacitance of a vacuum spherical capacitor from Gauss’s law. Explain the large-b limit.
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Compare with the answer and four-point rubric
- 1 point: Eᵣ4πr² = Q/ε₀ gives the gap field.
- 1 point: Integrate ∫ₐᵇEᵣdr to get ΔV = Q(1/a−1/b)/(4πε₀).
- 1 point: C = Q/ΔV = 4πε₀ab/(b−a).
- 1 point: As b→∞, b/(b−a)→1 and C→4πε₀a.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Why is E nonuniform here?
The same flux passes through growing spherical areas.
RECALL 2Which conductor is at higher V for positive inner Q?
The inner conductor.
RECALL 3Does Q appear in the final C?
No; it cancels for this fixed linear geometry.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Integrate the field between concentric spheres
- Eᵣ = Q/(4πε₀r²) in the gap.
- V(a) − V(b) = Q(1/a−1/b)/(4πε₀).
- C = 4πε₀ab/(b−a).
Remember: The outer radius is the shell’s inner surface radius. Use an integral because E is not uniform.
Conditions: Vacuum gap between concentric spheres with equal opposite charges. Outer radius b refers to the shell inner surface. Cross-section radii share a scale; field graph covers only the gap, not the metal.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (official Unit 10) · Objectives 10.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.3, objectives 10.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 3: Conductors and Capacitors, numbered Unit 10 in the official combined Physics C sequence. Topics 10.1–10.4 retain their official identifiers. Models state the electrostatic conditions, geometry approximations and whether charge or voltage stays fixed. Capacitor geometries include parallel plates, concentric spheres and long coaxial cylinders. Dielectric comparisons assume a fully filling ideal linear material. The optional 3D plate view uses explicitly different gap and lateral scales to show the small separation. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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